Question: Simplify; express your answer in exponential form. Assume $n\neq 0, a\neq 0$. $\dfrac{{(n^{-2}a)^{-4}}}{{(n^{3}a^{2})^{-1}}}$
Solution: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(n^{-2}a)^{-4} = (n^{-2})^{-4}(a)^{-4}}$ On the left, we have ${n^{-2}}$ to the exponent ${-4}$ . Now ${-2 \times -4 = 8}$ , so ${(n^{-2})^{-4} = n^{8}}$ Apply the ideas above to simplify the equation. $\dfrac{{(n^{-2}a)^{-4}}}{{(n^{3}a^{2})^{-1}}} = \dfrac{{n^{8}a^{-4}}}{{n^{-3}a^{-2}}}$ Break up the equation by variable and simplify. $\dfrac{{n^{8}a^{-4}}}{{n^{-3}a^{-2}}} = \dfrac{{n^{8}}}{{n^{-3}}} \cdot \dfrac{{a^{-4}}}{{a^{-2}}} = n^{{8} - {(-3)}} \cdot a^{{-4} - {(-2)}} = n^{11}a^{-2}$